Answer:
The condition necessary for the person to be able to remain stationary in mid-air is that there must be an upward force also referred to as THRUST
Explanation:
Thrust in Physics is defined as the force that propels, forces or pushes the mass of a certain object in a specified or particular direction.
Thrust can defined in the terms of a Jet pack can be defined as the force that is required to propel the mass of a person in an upward direction.
Based on the diagram that we have been shown the question, we can see that water is the fuel used to proper the user or person upwards hence, the jet pack used in the question is an HydroJet pack
The condition necessary for the person to be able to remain stationary in mid-air is that the propelling force called the thrust must be upwards such that it can overcome these 4 things:
a) Gravitational force
b) Mass of the jet pack user
c) Mass of the jet pack itself
d) The water which serves as fuel for the jet pack.
260 joules is hopefully right.
Answer:
Explanation:
If two forces act on an object in the same direction, the net force is equal to the sum of the two forces.
Answer:
The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
Explanation:
Given that,
Mass = 2.15 kg
Distance = 0.0895 m
Amplitude = 0.0235 m
We need to calculate the spring constant
Using newton's second law

Where, f = restoring force


Put the value into the formula


We need to calculate the kinetic energy of the mass
Using formula of kinetic energy

Here, 

Here, 


Put the value into the formula


Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.